Math help? Ball dropped from a height of 70 meters?
David drops a ball from a bridge at an initial height of 70 meters.
a.)What is the height of the ball to the nearest tenth of a meter exactly 2 seconds after he releases the ball?
b.)How many seconds after the ball is released will it hit the ground?
Please and Thank you. I don't understand how to solve it.
The position function for a free-falling object is given by
h(t) = - 4.88t2 + v0t + h0
where h = height in meters, t = time in secs., v0 = initial velocity in m./sec. and h0 = initial height in meters.
v0 = 0
h0 = 70 m
Position Function for this Problem:
h(t) = - 4.88t2 + 70
a.) t = 2 secs.
h(2) = - 4.88(2)2 + 70
h(2) = - 4.88(4) + 70
h(2) = -19.52 + 70
h(2) = 50.5
Height of Ball at 2 secs. = 50.5 meters
b.) Time to Impact: h(t) = 0:
- 4.88t2 + 70 = 0
- 4.88t2 = - 70
t2 = - 70 / - 4.88
t2 = 14.34
t = √14.24
t = 3.8
Time to Impact = 3.8 secs.
This is a physics problem. The ball is released with an initial velocity of 0 and accelerates at 9.81 m/s2 until it hits the ground. That should be enough to get you started.